## NCERT Solutions For Class 6 Maths Playing With Numbers Exercise 3.6

- Playing with Numbers Class 6 Ex 3.1
- Playing with Numbers Class 6 Ex 3.2
- Playing with Numbers Class 6 Ex 3.3
- Playing with Numbers Class 6 Ex 3.4
- Playing with Numbers Class 6 Ex 3.5
- Playing with Numbers Class 6 Ex 3.6
- Playing with Numbers Class 6 Ex 3.7
- Playing with Numbers Class 6 Extra Questions

**NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6**

**Exercise 3.6**

Question 1.

Find the HCF of the following numbers:

(a) 18, 48

(b) 30, 42

(c) 18, 60

(d) 27,63

(e) 36,84

(f) 34, 102

(g) 70, 105, 175

(h) 91, 112, 49

(i) 18, 54, 81

(j) 12, 45, 75

Solution:

(a) Given numbers are 18 and 48.

Prime factorisations of 18 and 48 are:

Here, the common factors are 2 and 3.

Hence, the HCF = 2 x 3 = 6.

(b) The given numbers are 30 and 42.

Prime factorisations of 30 and 42, are:

Here, the common factors are 2 and 3.

Hence, the HCF = 2 x 3 = 6.

(c) Given numbers are 18 and 60.

Prime factorisations of 18 and 60 are:

Here, the common factors are 2 and 3.

Hence, the HCF of 18 and 60 = 2 x 3 = 6.

(d) Given numbers are 27 and 63.

Prime factorisations of 27 and 63 are:

Here, the common factor is 3 (occurring twice).

Hence, the HCF = 3 x 3 = 9.

(e) Given numbers are 36 and 84.

Prime factorisations of 36 and 84 are:

Here, the common factors are 2, 2 and 3.

Hence, the HCF = 2 x 2 x 3 = 12.

(f) Given numbers are 34 and 102.

Prime factorisations of 34 and 102 are:

Here, the common factors are 2 and 17.

Thus, HCF is 2 x 17 = 34.

(g) The given numbers are 70, 105 and 175.

Prime factorisatios of 70, 105 and 175 are:

Here, common factors are 5 and 7.

Hence, the HCF of 70, 105 and 175 is 5 x 7 = 35.

(h) Given numbers are 91, 112 and 49.

Prime factorisations of 91, 112 and 49 are:

Here, the common factor is 7.

Hence, the HCF = 7.

(i) Given numbers are 18, 54 and 81.

Prime factorisations of 18, 54 and 81 are:

Here, the common factor is 3 (occurring twice).

Thus, the HCF = 3 x 3 = 9.

(j) Given numbers are 12, 45 and 75.

Prime factorisations of 12, 45 and 75 are:

Here, the common factor is 3.

Hence, the HCF = 3.

Question 2.

What is the HCF of two consecutive

(a) numbers?

(b) even numbers?

(c) odd numbers?

Solution:

(a) The common factor of two consecutive numbers is always 1.

Hence, the HCF = 1.

(b) The common factors of two consecutive even numbers are 1 and 2.

Hence, the HCF = 1 x 2 = 2.

(c) The common factor of two consecutive odd numbers is 1.

Hence, the HCF = 1.

Question 3.

HCF of co-prime numbers 4 and 15 was found as follows by factorisation:

4 = 2 x 2 and 15 = 3 x 15. Since there is no common prime factors, so HCF of 4 and 15 is 0.

Is the answer correct? If not, what is the correct HCF?

Solution:

No, answer is not correct.

Reason: 0 is not the prime factor of any number.

1 is always the prime factor of co-prime number.

Hence, the correct HCF of 4 and 15 is 1.

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